Timeline for Jackknife estimation of variance very different from expected variance
Current License: CC BY-SA 4.0
8 events
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| Aug 12, 2018 at 20:24 | comment | added | Sunfoil | And I think my statement that the Jackknife results would be different for non-linear functions was misleading. What I meant was non-linear functions of averages like $(\bar x)^4$. If the non-linear function is inside the average, e.g. $\overline{x^4}$, Jackknife doesn't make a difference. | |
| Aug 12, 2018 at 20:23 | comment | added | Sunfoil | @Mattz I have extended my answer quite a bit. Probably you don't need that any more, but maybe someone else finds it useful. I've experimented a little, and I think you're right. The bias is negligible in that case. The relevant point seems to be to take into account the correlation between $\overline{x^2}$ and $\overline{x^4}$. Also, thanks for the book tip! Is it the same as this Arxiv article by Young? | |
| Aug 12, 2018 at 20:21 | history | edited | Sunfoil | CC BY-SA 4.0 |
elaborated on when Jackknife makes a difference
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| Jul 22, 2018 at 15:54 | comment | added | Mattz | It goes under the name of Binder cumulant, it's something proportional to $\frac{\overline{x^4}}{ (\overline{x^2})^2}$ . x_i's are just measurements of the magnetization of a spin system at different montecarlo times by the way. In that case I've read that a computational algorithm for the error is the way to go because of the non-linearity. By the way, just for the record, I've found a quite good book explaining this stuff: "everything you wanted to know abut data analysis and fitting but were afraid to ask"-P.Young. The bias related to JK is said to be negligible most of the times | |
| Jul 22, 2018 at 15:34 | comment | added | Sunfoil | @Mattz I don't know how to calculate a Binder coefficient (do you have a link for that?), but I guess in general all bets are off and the results can be arbitrarily different. You can still compare it to results from other resampling methods like bootstrapping. By the way, does the Binder coefficient as you compute it depend only on the sample average, like f(x̅), or is it a function of all individual data points, f(x_1, x_2,..., x_N)? | |
| Jul 22, 2018 at 1:14 | vote | accept | Mattz | ||
| Jul 20, 2018 at 0:31 | comment | added | Mattz | found the bug. Anyway I didn't even know that the Jackknife is intended to be used for linear functions. Apart from my trivial example I was requested to use it for example to calculate a Binder coefficient, which is absolutely not linear. In that case it's normal to get a very different result between JKvariance and usual variance? | |
| Jul 19, 2018 at 22:47 | history | answered | Sunfoil | CC BY-SA 4.0 |